Answer:
(c) 16 m/s²
Explanation:
The position is ![r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%5B3.0%20%5Ctext%7B%20m%7D%20-%20%284.00%20%5Ctext%7B%20m%2Fs%7D%29t%5D%5Chat%7Bi%7D%20%2B%20%5B6.0%20%5Ctext%7Bm%7D%20-%20%288.00%20%5Ctext%7B%20m%2Fs%7D%5E2%20%29t%5E2%20%5D%5Chat%7Bj%7D)
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The velocity is the first time-derivative of <em>r(t).</em>
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The acceleration is the first time-derivative of the velocity.
Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,
Its magnitude is 16 m/s².
Unusual precipitation patterns
Explanation:
change 0.5 g to kg so 0.005kg then change 100 ml to m so 0.001m so density=mass over volume so from there you can continue
Answer:
option ( a ) is correct .
Explanation:
Escape velocity on the earth = √ ( 2 GM / R )
where G is universal gravitational constant , M is mass of the earth and R is radius .
V₀ = √ ( 2 GM / R )
escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth
Radius of the planet = 4 R
escape velocity of planet = √ ( 2 GM / 4R )
= .5 x √ ( 2 GM / R )
= .5 V₀
option ( a ) is correct .
When you square the "year" of each planet and divide it by the cube of its distance, or axis from the sun, the number would be the same for all the planets
